Let `P=[(2alpha),(5),(-3alpha^(2))] and Q=[(2l,-m,5n)]` are two matrices, where `l, m, n, alpha in R`, then the value of determinant PQ is equal to

Options

A. 0
B. `-1`
C. 2
D. not possible

Answer

To find the value of the determinant of the product of the matrices \( P \) and \( Q \), we can follow these steps:

### Step 1: Define the matrices
Let:
\[ P = \begin{bmatrix} 2\alpha \\ 5 \\ -3\alpha^2 \end{bmatrix}, \quad Q = \begin{bmatrix} 2l & -m & 5n \end{bmatrix} \]

### Step 2: Calculate the product \( PQ \)
The product \( PQ \) is calculated as follows:
\[ PQ = P \cdot Q = \begin{bmatrix} 2\alpha \\ 5 \\ -3\alpha^2 \end{bmatrix} \cdot \begin{bmatrix} 2l & -m & 5n \end{bmatrix} \]
This results in a \( 3 \times 3 \) matrix:
\[ PQ = \begin{bmatrix} 2\alpha \cdot 2l & 2\alpha \cdot (-m) & 2\alpha \cdot 5n \\ 5 \cdot 2l & 5 \cdot (-m) & 5 \cdot 5n \\ -3\alpha^2 \cdot 2l & -3\alpha^2 \cdot (-m) & -3\alpha^2 \cdot 5n \end{bmatrix} \]
Calculating each element:
\[ PQ = \begin{bmatrix} 4\alpha l & -2\alpha m & 10\alpha n \\ 10l & -5m & 25n \\ -6\alpha^2 l & 3\alpha^2 m & -15\alpha^2 n \end{bmatrix} \]

### Step 3: Analyze the matrix for determinant calculation
Now we have:
\[ PQ = \begin{bmatrix} 4\alpha l & -2\alpha m & 10\alpha n \\ 10l & -5m & 25n \\ -6\alpha^2 l & 3\alpha^2 m & -15\alpha^2 n \end{bmatrix} \]

### Step 4: Check for identical columns
Notice that the first and third columns can be expressed as:
- First column: \( 4\alpha l \), \( 10l \), \( -6\alpha^2 l \)
- Third column: \( 10\alpha n \), \( 25n \), \( -15\alpha^2 n \)

If we factor out common terms, we can see that:
- The first column can be factored to \( l \) and the third column can be factored to \( n \).

However, if we look closely, we can see that the first and third columns are not identical but they can be linearly dependent under certain conditions.

### Step 5: Conclusion about the determinant
Since the determinant of a matrix is zero if any two columns are identical or if any column can be expressed as a linear combination of the others, we can conclude that:
- The first and third columns are not identical but they can be dependent on the values of \( \alpha, l, m, n \).

Thus, we can conclude:
\[ \text{det}(PQ) = 0 \]

### Final Answer
The value of the determinant \( \text{det}(PQ) \) is:
\[ \boxed{0} \]